Mr. Bond has forgotten his n digit locker code

Hi niks18,
IMO, from A, we get that the length of code is 7 digits & digit 9 is not repeated in the code. So we can find out the total number of trials possible. As far as minimum number of trials is concerned, it could be anything between 1 & total combination possible.

HiBunuel, could you kindly shed some light on it.

Regards

OE

To solve this Permutation problem we need three values-
a) the value of n = no of digit in the locker code
b) the digit that appeared only once
c) whether the locker code was ODD or EVEN because permutation will get impacted based on even/odd combination as we have one number that can be used only once

Statement 1:
this implies that \(n=7\) because \(7*2=14>10\)

this implies that the digit that appeared only once was \(9\)=\(3^2\)

this implies that the code is ODD

Thus we got all the information we were looking for. Hence Sufficient. Note: this is a DS problem hence we don't need an exact answer here.

Statement 2:
From this we get the information that the code is ODD and 9 appears only once, but the value of \(n\) cannot be determined,
if \(k=1\), \(n=7\) & if \(k=2\), \(n=13\). Thus two values of n are possible. Hence Insufficient

Option A
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FYI - calculation for number of trials

There are two possible cases. The number \(9\) comes at the end, or it comes at position 4th, 5th, or 6th.

For the first case, the number would look like: \(853 - - - 9\) or \(847 - - - 9\)

In both these cases, the blanks can be occupied by any of the available \(9\) digits (0, 1, 2, ..., 8).

Thus, total possible numbers would be \(2 × (9 × 9 × 9) = 1458\).

For the second case, the number \(9\) can occupy any of the given position 4th, 5th, or 6th, and there shall be an odd number at position 7th.

Thus, the total number of ways shall be \(2[3(9 × 9 × 4)] = 1944\). Hence, answer is \(1458+1944=3402\).